Find a normal vector and tangent plane to the surface
$x^2+4y^2=z^2$ at $(3,2,5)$
So, this is what I have done:
$f(x,y,z) = x^2+4y^2-z^2$
Partial Differentiation gives:
$\nabla f(x,y,z) = (2x,8y,-2z)$
Substituting in:
$\nabla f(x,y,z) = (6,16,-10)$
My normal vector is
$\vec r(t) = (3,2,5) + t(6,16,-10)$
I was wondering if I am correct thus far? Also, from this, how do I get the tangent plane?
$\endgroup$ 12 Answers
$\begingroup$There is no need to write the equation of the line , $r(t)$, which passes through $(x_0,y_0,z_0)$ and is normal to the surface. $\nabla f(x_0,y_0,z_0)$ is a vector normal to the surface $f(x,y,z)=0$ at the point $(x_0,y_0,z_0)$. So the only things you needs to do are
1) Compute the unit normal by ${\bf{n}} = {{\nabla f} \over {\left\| {\nabla f} \right\|}}$ or any other vector parallel to that which can be $ \nabla f$ itself.
2) Write the equation of the plane passing through $(x_0,y_0,z_0)$
$$ {\nabla f(\bf{x_0})} \cdot ( {\bf{x}} - {\bf{x_0}} )= \bf{0} $$
3) In your problem, computation is as follows
$$ {\bf{x_0}} = {(3,2,5)}\\ {\nabla f(\bf{x_0})}=(6,16,-10) $$
and hence
$$\eqalign{ & 6\left( {x - 3} \right) + 16\left( {y - 2} \right) - 10\left( {z - 5} \right) = 0 \cr & 6x + 16y - 10z + \left( { - 18 - 32 + 50} \right) = 0 \cr & 6x + 16y - 10z = 0 \cr} $$
is the equation of the tangent plane.
$\endgroup$ 3 $\begingroup$Yes, you are correct. Known the normal vector $v$ at the a point $P=(x_P,y_P,z_P)$ of the surface this gives you the equation of the tangent plane in $P$ :
$$v \cdot (x-x_P,y-y_P,z-z_P)=0.$$
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